(a)(i) State the laws of refraction of light. (ii) Describe an experiment to determine the refractive index, n, of the material of an equilaleral triangular...

Answer Details

(a)(i) Laws of refraction of light

• The incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane.
• The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair of media (Snell's law): \( \dfrac{\sin i}{\sin r} = n \).

(a)(ii) Determining the refractive index by the minimum-deviation method

Place the equilateral triangular glass prism flat on a sheet of paper and trace its outline. Remove the prism, and on one refracting face mark a point; draw the normal at this point and rule an incident ray meeting the face at a chosen angle of incidence \(i\). Replace the prism exactly on its outline. Stick two pins \(P_1\) and \(P_2\) vertically on the incident ray. Looking through the second refracting face, stick two more pins \(P_3\) and \(P_4\) so that they appear in a straight line with the images of \(P_1\) and \(P_2\). Remove the prism, ring the pin holes, and join \(P_3P_4\) to the face to give the emergent ray. Produce the incident and emergent rays forward until they meet; the angle between them is the angle of deviation \(d\). Measure \(i\) and \(d\) with a protractor.

Repeat the whole procedure for at least five values of \(i\), then plot a graph of the deviation \(d\) (vertical axis) against the angle of incidence \(i\) (horizontal axis). The graph is a smooth U-shaped curve; the deviation falls to a lowest value and then rises again. The value of \(d\) at the lowest point of the curve is the minimum deviation \(D\), read here as \(D = 37.2^\circ\).

With the refracting (apex) angle of the equilateral prism \(A = 60^\circ\), the refractive index is obtained from

\[ n = \frac{\sin\!\left(\dfrac{A + D}{2}\right)}{\sin\!\left(\dfrac{A}{2}\right)} = \frac{\sin\!\left(\dfrac{60^\circ + 37.2^\circ}{2}\right)}{\sin 30^\circ} = \frac{\sin 48.6^\circ}{\sin 30^\circ} = \frac{0.750}{0.500} = 1.50. \]

Precautions: keep pin traces neat and thin; fix all pins truly vertical; and avoid parallax when reading the protractor.

(b)(i) Ray diagram showing the apparent position of the mark

The mark \(O\) is on the paper directly beneath the glass. Rays leaving \(O\) strike the top surface and are refracted away from the normal as they pass from glass to air, so they diverge more steeply. To the eye above, these emergent rays appear to come from a point \(I\) which lies vertically above \(O\) but nearer the top surface. Thus the mark is seen raised: \(I\) is its apparent (virtual) position.

(b)(ii) Apparent displacement of the mark

For viewing normally through a parallel-sided block,

\[ n = \frac{\text{real depth}}{\text{apparent depth}} \quad\Rightarrow\quad \text{apparent depth} = \frac{\text{real depth}}{n} = \frac{12}{1.5} = 8\ \text{cm}. \]

Therefore the apparent displacement (the amount by which the mark appears raised) is

\[ \text{apparent displacement} = \text{real depth} - \text{apparent depth} = 12 - 8 = 4\ \text{cm}. \]